A mass is suspended from a wire of negligible mass. The length of the wire is and its cross-sectional area is If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency If

Question

A $0.1 kg$ mass is suspended from a wire of negligible mass. The length of the wire is $1 m$ and its cross-sectional area is $4.9 \times 10^{- 7} m^{2} .$ If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140 {rad s}^{- 1} .$ If the Young's modulus of the material of the wire is $n \times 10^{9} {Nm}^{- 2},$ the value of $n$ is ___. [2010]

Smart Achievers · Accepted Answer

(4)

$As we know,$

$Y = \frac{F L}{A l} \Rightarrow F = (\frac{Y A}{L}) l$

Also, $F = K l$

or, $K l = (\frac{Y A}{L}) l \Rightarrow K = \frac{Y A}{L}$

Angular frequency $ω = \sqrt{\frac{K}{m}}$ or, $ω = \sqrt{\frac{Y A}{m L}}$

or, $140 = \sqrt{\frac{n \times 10^{9} \times 4.9 \times 10^{- 7}}{0.1 \times 1}}$ $(∵ Y = 9 \times 10^{9} given)$

$∴$ $n = 4$

Solution

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Solution

Ans. (4) As we know, Y=FLAl⇒F=(YAL)l Also, F=Kl or, Kl=(YAL)l⇒K=YAL Angular frequency ω=Km or, ω=YAmL or, 140=n×109×4.9×10-70.1×1 (∵Y=9×109 given) ∴ n=4

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